# Find limits of sequences

Prove that

$$a) \lim\limits_{n \to \infty} (\frac{1^p+2^p+...+n^p}{n^p} - \frac{n}{p+1})=\frac{1}{2},$$

$$b) \lim\limits_{n \to \infty} \frac{1^p+3^p+...+(2n-1)^p}{n^{p+1}}=\frac{2^p}{p+1},$$

where is $p \in \Bbb N$.

Thanks to Stolz–Cesàro theorem in $a)$ I went to $$\lim\limits_{n \to \infty} \frac{(n+1)^p}{(n+1)^p-n^p}$$ which after dividing by $n^p$ goes to

$$\lim\limits_{n \to \infty} \frac{1+\frac{p-1}{n}+\frac{p(p-1)}{2n^2}+...+\frac{1}{n^p}}{\frac{p-1}{n}+\frac{p(p-1)}{2n^2}+...+\frac{1}{n^p}} = \frac{1}{\infty}$$ At the same time $$\lim\limits_{n \to \infty} \frac{n}{p+1}=\infty,$$

so initial limit should be $\frac{1}{\infty}-\infty = ?$

I can't figure out what I'm missing ( $b)$ seems to complicated to me so I didn't even try it).

• You are not new here..you know how it goes..what have you tried? Sep 18, 2017 at 19:43
• Faulhaber${{}}$? Sep 18, 2017 at 19:49
• Actually (b) is simpler than (a)...
– Did
Sep 18, 2017 at 20:06

We know that the sum of the $p^{th}$ powers of the integers is a polynomial of degree $p+1$ in $n$, let $S(n):=\alpha n^{p+1}+\beta n^p+R(n)$.

Then

$$n^p=S(n)-S(n-1)=\alpha n^{p+1}-\alpha (n-1)^{p+1}+\beta n^p-\beta (n-1)^p+R(n)-R(n-1)\\ =\alpha (p+1)n^p-\alpha \frac{(p+1)p}2n^{p-1}+\beta pn^{p-1}+Q(n)$$ where $Q$ is of degree at most $p-2$.

By identification,

$$\alpha =\frac1{p+1},\\\beta =\frac12$$ (which are the first two Faulhaber coefficients.)

This justifies the $\dfrac12$ in a).

For b), consider $$S(2n)-2^pS(n)=\frac{2^{p+1}n^{p+1}-2^pn^{p+1}}{p+1}+Q'(n)=\frac{2^pn^{p+1}}{p+1}+Q'(n)$$ to get the sum of the powers of the odd integers.

• Please correct it's not $\beta pn^{n-1}$ but surely $\beta pn^{p-1}$ that you meant in second line or am I wrong ? Sep 18, 2017 at 20:39
• @Isham: absolutely, thanks.
– user65203
Sep 18, 2017 at 20:40
• you're welcome..........nice job + 1 Sep 18, 2017 at 20:40

Solution for (b):

$$\lim_n\frac{1^p+3^p+\cdots +(2n-1)^p}{n^{p+1}}=\lim_n \frac{1^p+2^p+\cdots +(2n)^p}{n^{p+1}}-\lim_n\frac{2^p+4^p+\cdots+(2n)^p}{n^{p+1}}$$

$$=2^{p+1}\lim_n\frac{1^p+2^p+\cdots +n^p}{n^{p+1}}-2^p\lim_n\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}$$

$$=(2^{p+1}-2^p).\lim_n\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=2^p.\lim_n\frac{1}{n}\sum_{r=1}^n \left(\frac{r}{n}\right)^p$$

$$=2^p.\int_0^1x^p\,dx=\frac{2^p}{p+1}$$

• This answer would be more helpful (and I'd upvote it) if it were more detailed. Do we answer questions to show how smart we are, or do we want to help the OP?
– user436658
Sep 18, 2017 at 20:16
• How do we proceed from here? Sep 18, 2017 at 20:36
• @ProfessorVector Edited..! Sep 18, 2017 at 20:53
• @mechanodroid Please See the edition.! Sep 18, 2017 at 20:53
• Ok, I've promised. Why didn't you follow the original idea, a Riemann sum for $$\int^1_0(2x)^p\,dx?$$
– user436658
Sep 18, 2017 at 21:03

As you mentioned with Stolz–Cesàro theorem we will get

$$b)\lim _{ n\to \infty } \frac { { x }_{ n+1 }-{ x }_{ n } }{ { y }_{ n+1 }-{ y }_{ n } } =\lim _{ n\to \infty } \frac { 1^{ p }+3^{ p }+...+(2n+1)^{ p }-\left[ 1^{ p }+3^{ p }+...+(2n-1)^{ p } \right] }{ \left( n+1 \right) ^{ p+1 }-{ n }^{ p+1 } } =\\ =\lim _{ n\to \infty } \frac { { \left( 2n+1 \right) }^{ p } }{ { \left( p+1 \right) n }^{ p }+...+1 } =\lim _{ n\to \infty } \frac { { \left( 2n \right) }^{ p }+...+1 }{ { \left( p+1 \right) n }^{ p }+...+1 } =\frac { { 2 }^{ p } }{ p+1 }$$